Q: A uniform chain of length l, total mass m and containing many links, is held at one end over a table with the other end just touching the table top.
a)The chain is released and falls freely. What is the speed of the falling section of the chain at time t after release? What is now the force between the links?
b) Work out the increment of mass , which hits the table in the increment of time . Find the corresponding change in momentum and hence the instantaneous force on the table.
c) What is the total force acting on the table as a function of time? Show that the maximum value of the total force is three times the total weight of the chain.

I have no idea as to how to approach the question, other than for part a) the speed of the falling section of chain is: v=u+at, u=0 and v=at, but don't know if that's right.

Any help would be hugely appreciated.

Oct 19th 2008, 07:25 AM

HallsofIvy

Quote:

Originally Posted by free_to_fly

I'm currently stuck on this question:

Q: A uniform chain of length l, total mass m and containing many links, is held at one end over a table with the other end just touching the table top.
a)The chain is released and falls freely. What is the speed of the falling section of the chain at time t after release? What is now the force between the links?
b) Work out the increment of mass , which hits the table in the increment of time . Find the corresponding change in momentum and hence the instantaneous force on the table.
c) What is the total force acting on the table as a function of time? Show that the maximum value of the total force is three times the total weight of the chain.

I have no idea as to how to approach the question, other than for part a) the speed of the falling section of chain is: v=u+at, u=0 and v=at, but don't know if that's right.

Any help would be hugely appreciated.

So you are holding the upper end of the chain a distance l above the table?
It's probably simplest to do (a) by using "conservation of energy". Taking the table top as 0 potential energy, the initial potential energy of the chain is the integral, from 0 to l, of (mg)/l x dx which is (1/2)mgl (the potential energy if the chain had all been at its midpoint- that's reasonable!). If, at given time t, the upper end of the chain has fallen fallen a distance y, then all but l-y is now on the table and so has 0 potential energy. The remaining, still falling, part of the chain has potential energy (1/2)(mg/l)(l-y)^2 by the same integration. The kinetic energy of the chain must be the difference, (1/2)(mg/l)(2yl- y^2) and that must be equal to 1/2 mv^2= 1/2 m (dy/dt)^2. y must satisfy the differential equation dy/dt= sqrt((mg/l)(2yl- y^2). And, of course, dy/dt is the speed asked for in part (a).